For example, if I have a gaussian surface and a negative charge somewhere outside that gaussian surface, there's going to be a non-zero electric flux.

However, According to gauss's law, electric flux = $\frac{Q}{E}$ and $Q$ is $0$ because there are no charges inside the surface?

  • $\begingroup$ The flux is zero. It's positive in some places and negative in others. $\endgroup$
    – knzhou
    Commented Jan 17, 2016 at 3:44
  • $\begingroup$ What about the electric field? That's not 0 $\endgroup$ Commented Jan 17, 2016 at 3:57

1 Answer 1


As the charge is outside the surface the flux entering the surface is equal to the flux leaving the surface and total flux is therefore zero.

The electric field is not zero. Flux =$\int \mathbf{E.ds}$ is zero not the electric field itself. for a negative single charge electric filed is $\mathbf{E}=\frac{Q}{4\pi \epsilon_0 r^2}(-\hat{r})$. The flux is zero because $\mathbf{ds}$ has different direction with respect to $\hat{r}$ at different point on the surface and the total flux is therefore zero.


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